Solved Examples based on GP

Solved examples :

The 7thterm of a G.P. is 8 times the 4thterm. Find the G.P. when its 5thterm is 48.

Solution:

Given that t7= 8t4⇒ ar6= 8ar3

⇒r3= 8 = 23⇒r = 2.

Also t5= 48 Þ ar4= 48 or 16a = 48⇒a = 3.

Hence the required G.P. is 3, 6, 12, 24 ……

Illustration:

Does there exists a G.P. containing 27, 8 and 12 as three of its terms? If it exists, how many such progressions are possible?

Solution:

Let 8 be the mth, 12 the nthand 27 be the tthterms of a G.P. whose first term is A and common ratio is R.

Then 8 = ARm–1, 12 = ARn–1, 27 = ARt–1

⇒8/12 = Rm–n= 2/3, 12/27 = Rn–t= (2/3)2, 8/27 = Rm–t=(2/3)3

⇒2m – 2n = n – t and 3m – 3n = m – t

⇒2m + t = 3n and 2m + t = 3n

⇒2m+t/3 = n.

There are infinite sets of values of m, n, t which satisfy this relation. For example, take m = 1, then 2+t/3 = n = k⇒n = k, t = 3k – 2. By giving different values to k we get integral values of n and t. Hence there are infinite numbers of G.P.’s whose terms may be 27, 8, 12 (not consecutive).

Illustration:

In a four term series if first three are in G.P. and last three are in A.P. with common different 6 and last terms is equal to the first term then find all four terms in series.

Solution:

This is very tricky question. If you read question carefully then it is clear that we have to start with A.P. because common difference is given.

Let the numbers be a + 6, a–6, a, a+6 now first three are in G.P. is (a–6)2= a(a+6) or, a2– 12a + 36 = a2therefore numbers are 8, –4, 2, 8.